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Question Number 136750 by mnjuly1970 last updated on 25/Mar/21

$$.....advancedcalculus.....$$$$provethat::$$$$...\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{1-e^{-x^2}}{x^2}dx=\sqrt{\pi }$$

Answered by Ñï= last updated on 25/Mar/21

$${\int }_0^{\mathrm{\infty }}\frac{1-e^{-x^2}}{x^2}dx={\int }_0^{\mathrm{\infty }}\frac{e^{-0}-e^{-x^2}}{x^2}dx={\int }_0^{\mathrm{\infty }}{\int }_0^1{e}^{-{ax}^2}dadx$$$$={\int }_0^1\frac{\sqrt{\pi }}{2\sqrt{a}}da=\sqrt{\pi }$$

Commented by mnjuly1970 last updated on 25/Mar/21

$$thanksalot...$$

Answered by Dwaipayan Shikari last updated on 25/Mar/21

$$\eta \left(\varphi \right)={\int }_0^{\mathrm{\infty }}\frac{e^{-\varphi x^2}-e^{-x^2}}{x^2}dx\Rightarrow {\eta }^{\prime }\left(\varphi \right)=-{\int }_0^{\mathrm{\infty }}{e}^{-\varphi x^2}dx=-\frac{\sqrt{\pi }}{2\sqrt{\varphi }}$$$$\eta \left(\varphi \right)=-\sqrt{\pi \varphi }+C\Rightarrow \eta \left(1\right)=0\Rightarrow C=\sqrt{\pi }$$$$\eta \left(\varphi \right)=\sqrt{\pi }(1-\sqrt{\varphi })\Rightarrow \eta \left(0\right)=\sqrt{\pi }$$

Answered by mathmax by abdo last updated on 25/Mar/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{ax}}^2}-{\mathrm{e}}^{-{\mathrm{x}}^2}}{{\mathrm{x}}^2}\mathrm{dx}\mathrm{with}\mathrm{a}>0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ax}}^2}\mathrm{dx}$$$$=-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(\sqrt{\mathrm{a}}\mathrm{x}\right)}^2}\mathrm{dx}{=}_{\sqrt{\mathrm{a}}\mathrm{x}=\mathrm{t}}-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\frac{\mathrm{dt}}{\sqrt{\mathrm{a}}}=-\frac{1}{\sqrt{\mathrm{a}}}.\frac{\sqrt{\pi }}{2}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\sqrt{\pi }\int \frac{\mathrm{da}}{2\sqrt{\mathrm{a}}}+\mathrm{C}$$$$=-\sqrt{\pi \mathrm{a}}+\mathrm{C}$$$$\mathrm{f}\left(1\right)=0=-\sqrt{\pi }+\mathrm{C}\Rightarrow \mathrm{C}=\sqrt{\pi }\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\sqrt{\pi }-\sqrt{\pi \mathrm{a}}$$$$\mathrm{\Phi }=\mathrm{f}\left(0\right)=\sqrt{\pi }$$

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